HDU 3172 (STL map)

Virtual Friends

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4665    Accepted Submission(s): 1320

Problem Description

These days, you can do all sorts of things online. For example, you can use various websites to make virtual friends. For some people, growing their social network (their friends, their friends' friends, their friends' friends' friends, and so on), has become an addictive hobby. Just as some people collect stamps, other people collect virtual friends.

Your task is to observe the interactions on such a website and keep track of the size of each person's network.

Assume that every friendship is mutual. If Fred is Barney's friend, then Barney is also Fred's friend.

Input

Input file contains multiple test cases.
The first line of each case indicates the number of test friendship nest.
each friendship nest begins with a line containing an integer F, the number of friendships formed in this frindship nest, which is no more than 100 000. Each of the following F lines contains the names of two people who have just become friends, separated by a space. A name is a string of 1 to 20 letters (uppercase or lowercase).

Output

Whenever a friendship is formed, print a line containing one integer, the number of people in the social network of the two people who have just become friends.

Sample Input

   
   
   
   

    
    
    
    1
3
Fred Barney
Barney Betty
Betty Wilma
   
   
   
   

Sample Output

   
   
   
   

    
    
    
    2
3
4
   
   
   
   

与上篇我写的并查集差不多，只是这里需要用到map容器了，学习到了map容器中 .find() 的用法，↖(^ω^)↗

#include<cstdio>
#include<string>
#include<map>
using namespace std;

const int num = 100001;
int p[num], sum[num];
char s1[num][21], s2[num][21];

int find(int x) {return p[x] == x ? x : p[x] = find(p[x]);}

int main()
{
    int T, n, i;

    while(~scanf("%d", &T))
    {
        while(T--)
        {
            scanf("%d", &n);
            for(i=1; i<=num; i++) p[i] = i, sum[i] = 1;
            map<string,int> m;
            int g = 1; //人物个数
            for(i=1; i<=n; i++)
            {
                scanf("%s%s", s1[i],s2[i]);
                if(m.find(s1[i]) == m.end())  //说明：遍历整个map容器，判断键值x是否在map中存在。
                    m[s1[i]] = g++;           //参数：x。待搜索的键值，注意，其实是指针！
                if(m.find(s2[i]) == m.end())  //返回：成功返回指向键值为x的元素的迭代器，失败返回map.end().
                    m[s2[i]] = g++;
                int x = find(m[s1[i]]);
                int y = find(m[s2[i]]);
                if(x != y)
                {
                    p[x] = y;
                    sum[y] += sum[x];
                }
                printf("%d\n", sum[y]);
            }
        }
    }
    return 0;
}